At what temperature would the root-mean-square speed of hydrogen,H₂,molecules equal 11.2 km/s (the earth's escape speed)? The mass of a hydrogen atom is 1.67 × 10^-27kg,and the Boltzmann constant is 1.38 × 10^-23 J/K. A)1.01 × 10² K B)1.01 × 10⁴ K C)1.01 × 10⁶ K D)1.01 × 10⁸ K

Question

Quizplus · Accepted Answer

The root-mean-square speed, $v_{rms}$, of a gas molecule can be calculated using the formula $v_{rms} = \sqrt{\frac{3kT}{m}}$, where $k$ is the Boltzmann constant, $T$ is the temperature in Kelvin, and $m$ is the mass of a gas molecule. Rearranging the formula to solve for $T$, we get $T = \frac{m(v_{rms})^2}{3k}$. Substituting the given values: $m = 1.67 \times 10^{-27} \, \text{kg}$, $v_{rms} = 11.2 \times 10^3 \, \text{m/s}$ (since 11.2 km/s is equal to 11.2 \times 10^3 m/s), and $k = 1.38 \times 10^{-23} \, \text{J/K}$, we find that $T$ calculates to a value in the order of $10^{14}$ K, which corresponds to choice B.

At What Temperature Would the Root-Mean-Square Speed of Hydrogen,H2,molecules Equal

At What Temperature Would the Root-Mean-Square Speed of Hydrogen,H₂,molecules Equal